BE FIRST TO ANSWER FOR SUM GOOD!!!!!:D

Which statement is true of chloroplasts?

Anastasy [175]

d answer is correct that help

5 0

3 years ago

A buret was improperly read to 1 decimal place giving a reading of 27.2 mL. If the actual volume is 27.26 mL, what is the percen

Nezavi [6.7K]

Percent error can be calculated by the difference of the theoretical value and the measured value divided by the theoretical value multiplied by 100 percent.

% error = 27.26 - 27.2 / 27.26 x100
% error = 0.22%

A value close to zero would mean that the measured value is more or less near the actual value.

5 0

3 years ago

Read 2 more answers

An electrochemical cell has the following standard cell notation.

icang [17]

The cell notation is:

$Mg(s)|Mg^{+2}(aq)||Ag^{+}(aq)|Ag(s)$

here in cell notation the left side represent the anodic half cell where right side represents the cathodic half cell

in anodic half cell : oxidation takes place [loss of electrons]

in cathodic half cell: reduction takes place [gain of electrons]

1) this is a galvanic cell

2) the standard potential of cell will be obtained by subtracting the standard reduction potential of anode from cathode

$E^{0}_{Mg}=-2.38V$

$E^{0}_{Ag}=+0.80V$

Therefore

$E^{0}_{cell}=0.80-(-2.38)=+3.18V$

3) as the value of emf is positive the reaction will be spontaneous as the free energy change of reaction will be negative

Δ $G^{0}=-nFE^{0}$

As reaction is spontaneous and there will be conversion of chemical energy to electrical energy it is a galvanic cell.

7 0

3 years ago

Isotopes of an element have the same number of ________ and different numbers of ________. neutrons... protons protons... neutro

Ivahew [28]

Protons . . . neutrons

8 0

3 years ago

Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

$A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$ = 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0

3 years ago